Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Every separable metric space is second countable.

Short Answer

Expert verified
A separable metric space has a countable, dense subset, using which a countable basis of rational open balls can be constructed.

Step by step solution

01

Understanding the Problem

First, we need to understand the terms: a 'separable metric space' means that there exists a countable, dense subset within the metric space. A 'second countable space' means that the space has a countable base for its topology. The problem asks us to show that if one exists, then so does the other.
02

Identify the Known Elements

Given that the metric space is separable, there exists a countable dense subset, say \( D = \{ d_1, d_2, d_3, \ldots \} \), where every point in the space can be approximated as closely as desired by points from \( D \).
03

Construct a Basis

To show the space is second countable, we need to construct a countable basis. Consider the set of open balls of the form \( B(d_i, r) \) where \( d_i \) is in \( D \) and \( r \) is a rational number. Rational numbers are countable, thus making the collection of these balls countable.
04

Verify Basis Coverage

We need to verify that the collection of open balls covers the space. Given any point \( x \) in the metric space and \( \epsilon > 0 \), there exists an element \( d_i \) in \( D \) such that \( d(x, d_i) < \epsilon \). Thus, \( x \) is in the open ball \( B(d_i, \epsilon) \), demonstrating that every point in the space can be captured by some ball in our constructed set.
05

Verify Basis Conditions

A basis should satisfy the condition that any open set within the metric space can be expressed as a union of basis elements. Since \( D \) is dense and rational balls are countable, any open set can be expressed as a union of these balls, meeting the condition for a basis.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Second Countability
Second countability is a concept in topology that refers to the existence of a countable basis for a topology. In simpler terms, a space is second countable if there is a set of open sets (called a basis) such that any open set in the space can be formed by taking unions of these basis elements. What's important here is the 'countable' part, meaning that these basis elements can be listed as a sequence like a list of integers. Now, why does second countability matter? One reason is it links to separability. A separable space has a countable dense subset, and if a metric space is second countable, it implies that the space is well-behaved by allowing sequences convergence to describe its topology. Second countable spaces have nice properties:
  • They are Lindelöf spaces, meaning every open cover has a countable subcover.
  • They are also separable, as shown in our exercise, making them accessible for analysis.
Understanding and constructing the basis for second countable spaces often revolves around clever selections of dense subsets and using aspects like rational numbers, which are naturally countable, to construct these bases.
Metric Space Topology
Metric space topology refers to the study of topological properties of metric spaces, which are sets equipped with a metric. A metric is a way to define the distance between any two points in a space. Imagine a surface, and the metric lets you measure the distance between any two points on it. A metric space provides the framework where this concept of distance makes sense, and allows us to then discuss concepts such as open and closed sets, diameter, and convergent sequences. A topology on a metric space is defined by its open sets, with the most common type being open balls. An open ball centered at a point with a radius is the set of all points that are within that distance (radius) from the center. This gives rise to intuitive concepts like neighborhoods around points. A metric space's topology guides how we understand continuity and convergence:
  • An open set is a collection of points forming a space where each point has neighbors close enough to not "leave" the set.
  • Closed sets, conversely, include all their boundary points, without any points "missing" at the edges.
Metric space topology provides the foundational framework and language to talk about convergence, compactness, and continuity in a mathematically rigorous way.
Dense Subsets
Dense subsets in topology are crucial as they provide the groundwork for many concepts like separability and second countability. A dense subset of a space means that every point in the space is either in the dense subset or can be arbitrarily closely approximated by points from the subset. In a metric space, if you think of this concept visually, a dense subset is like a dotted pattern spread over a canvas, where every spot on the canvas merges into the pattern when viewed up close. For a set to be dense, any open set in the space must intersect with the dense subset at some point, however small the open set is. Why are dense subsets favored in topology? Utilization of dense subsets allows:
  • Efficient approximations of anything within the space, facilitating calculations and constructions.
  • The development of other important properties, such as separability — if a space has a countable dense subset, it is called separable.
Dense subsets help in constructing bases for metric spaces, leading us back to pivotal concepts like second countability. Knowing there's a countable dense subset helps pick countably many open covers that ensure every open set in the space can be represented as a union of these.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A topological space \(X\) is called disconnected if there exist nonempty open sets \(U, V\) such that \(U \cap V=\varnothing\) and \(U \cup V=X\); otherwise \(X\) is connected. When we speak of connected or disconnected subsets of \(X\), we refer to the relative topology on them. a. \(X\) is connected iff \(\varnothing\) and \(X\) are the only subsets of \(X\) that are both open and closed. b. If \(\left\\{E_{\alpha}\right\\}_{\alpha \in A}\) is a collection of connected subsets of \(X\) such that \(\bigcap_{\alpha \in A} E_{\alpha} \neq\) \(\varnothing\), then \(\bigcup_{\alpha \in A} E_{\alpha}\) is connected. c. If \(A \subset X\) is connected, then \(\bar{A}\) is connected. d. Every point \(x \in X\) is contained in a unique maximal connected subset of \(X\), and this subset is closed. (It is called the connected component of \(x\).)

Let \(X\) and \(Y\) be compact Hausdorff spaces. The algebra generated by functions of the form \(f(x, y)=g(x) h(y)\), where \(g \in C(X)\) and \(h \in C(Y)\), is dense in \(C(X \times Y)\).

Suppose that \((X, \mathcal{J})\) is a compact Hausdorff space and \(\mathcal{T}^{\prime}\) is another topology on \(X .\) If \({ }^{\prime}{ }^{\prime}\) is strictly stronger than \({\mathcal{J}}\), then \(\left(X, \mathcal{J}^{\prime}\right)\) is Hausdorff but not compact. If \(\mathcal{J}^{\prime}\) is strictly weaker than \(\mathcal{J}\), then \(\left(X, \mathcal{T}^{\prime}\right)\) is compact but not Hausdorff.

Let \(X\) be a topological space, \(Y\) a Hausdorff space, and \(f, g\) continuous maps from \(X\) to \(Y\). a. \(\\{x: f(x)=g(x)\\}\) is closed. b. If \(f=g\) on a dense subset of \(X\), then \(f=g\) on all of \(X\).

Let \(0^{\prime}\) denote a point that is is not an element of \((-1,1)\), and let \(X=(-1,1) \cup\) \(\left\\{0^{\prime}\right\\}\). Let \(\mathcal{T}\) be the topology on \(X\) generated by the sets \((-1, a),(a, 1),[(-1, b) \backslash\) \(\\{0\\}] \cup\left\\{0^{\prime}\right\\}\), and \([(c, 1) \backslash\\{0\\}] \cup\left\\{0^{\prime}\right\\}\) where \(-1

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free